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Question Number 66743 by pete last updated on 19/Aug/19

Each month a store owner can spend at most $100,000 on PC′s and laptops. A PC costs the store owner $1000 and a laptop costs him $1500. Each PC is sold for a profit of $400 while a laptop is sold for a profit of $700. The store owner estimates that at least 15 PC′s but no more than 80 are sold each month. He also estimates that the number of laptops sold is at most half the PC′s. How many PC′s and how many laptops should be sold in order to maximize the profit?

$$\mathrm{Each}\:\mathrm{month}\:\mathrm{a}\:\mathrm{store}\:\mathrm{owner}\:\mathrm{can}\:\mathrm{spend}\:\mathrm{at} \\ $$$$\mathrm{most}\:\$\mathrm{100},\mathrm{000}\:\mathrm{on}\:\mathrm{PC}'\mathrm{s}\:\mathrm{and}\:\mathrm{laptops}.\:\mathrm{A} \\ $$$$\mathrm{PC}\:\mathrm{costs}\:\mathrm{the}\:\mathrm{store}\:\mathrm{owner}\:\$\mathrm{1000}\:\mathrm{and}\:\mathrm{a} \\ $$$$\mathrm{laptop}\:\mathrm{costs}\:\mathrm{him}\:\$\mathrm{1500}.\:\mathrm{Each}\:\mathrm{PC}\:\mathrm{is}\:\mathrm{sold} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{profit}\:\mathrm{of}\:\$\mathrm{400}\:\mathrm{while}\:\mathrm{a}\:\mathrm{laptop}\:\mathrm{is}\:\mathrm{sold} \\ $$$$\mathrm{for}\:\mathrm{a}\:\mathrm{profit}\:\mathrm{of}\:\$\mathrm{700}.\:\mathrm{The}\:\mathrm{store}\:\mathrm{owner}\:\mathrm{estimates} \\ $$$$\mathrm{that}\:\mathrm{at}\:\mathrm{least}\:\mathrm{15}\:\mathrm{PC}'\mathrm{s}\:\mathrm{but}\:\mathrm{no}\:\mathrm{more}\:\mathrm{than} \\ $$$$\mathrm{80}\:\mathrm{are}\:\mathrm{sold}\:\mathrm{each}\:\mathrm{month}.\:\mathrm{He}\:\mathrm{also}\:\mathrm{estimates} \\ $$$$\mathrm{that}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{laptops}\:\mathrm{sold}\:\mathrm{is}\:\mathrm{at}\:\mathrm{most} \\ $$$$\mathrm{half}\:\mathrm{the}\:\mathrm{PC}'\mathrm{s}.\:\mathrm{How}\:\mathrm{many}\:\mathrm{PC}'\mathrm{s}\:\mathrm{and}\:\mathrm{how} \\ $$$$\mathrm{many}\:\mathrm{laptops}\:\mathrm{should}\:\mathrm{be}\:\mathrm{sold}\:\mathrm{in}\:\mathrm{order}\:\mathrm{to} \\ $$$$\mathrm{maximize}\:\mathrm{the}\:\mathrm{profit}? \\ $$

Commented by Prithwish sen last updated on 19/Aug/19

let no. of PC sold be x and no. of laptop sold be y 1000x+1500y≤100000 15≤x≤80 and 0≤y≤40 subject to maximize 400x+700y I think it is easy to solve using graphical method.

$$\mathrm{let}\:\mathrm{no}.\:\mathrm{of}\:\mathrm{PC}\:\mathrm{sold}\:\mathrm{be}\:\mathrm{x}\:\mathrm{and}\:\mathrm{no}.\:\mathrm{of}\:\mathrm{laptop}\:\mathrm{sold}\:\mathrm{be}\:\mathrm{y} \\ $$$$\mathrm{1000x}+\mathrm{1500y}\leqslant\mathrm{100000} \\ $$$$\mathrm{15}\leqslant\mathrm{x}\leqslant\mathrm{80}\:\mathrm{and}\:\mathrm{0}\leqslant\mathrm{y}\leqslant\mathrm{40} \\ $$$$\mathrm{subject}\:\mathrm{to}\:\mathrm{maximize} \\ $$$$\mathrm{400x}+\mathrm{700y} \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{is}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{using}\:\mathrm{graphical}\:\mathrm{method}. \\ $$

Commented by Tony Lin last updated on 20/Aug/19

Commented by Tony Lin last updated on 20/Aug/19

400×40+700×40=44000

$$\mathrm{400}×\mathrm{40}+\mathrm{700}×\mathrm{40}=\mathrm{44000} \\ $$

Commented by pete last updated on 20/Aug/19

Thank you sir, very much

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir},\:\mathrm{very}\:\mathrm{much} \\ $$

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